Question

What does it mean for a number to be weighted?

For example, 8421 BCD codes are weighted and XS3 codes are not. (I only know this because I am regurgitating what I read but, I don't grasp it.)

According to some random stuff I found on the Internet, weights have to do with the position of digits and that shifting the position of those digits matters however, I can't picture a case where shifting the position of digits wouldn't matter.

Sorry for being pedantic about something that's probably not so important but, I just want to know!

Answer 1

Binary Coded Decimal (BCD) is weighted (like our decimal system) in that each digit has a certain weight assigned to it. The number can be recovered by simple summing the right weights.

Excess-3 (XS3) (and Gray code being another example) is not weighted. For Gray code, it is more clear. You cannot say that a certain bit always will have a value of 2. While in XS3 each bit has a weight (like in BCD), you cannot simply sum all the weights. You have to add +3 to it. Therefor, it is not weighted.


An example (BCD): the value of \[0110 = 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 = 6\]

The same number in XS3 would be \[0110 = 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 + 3 = 9\]

The extra +3 you need to use make XS3 non-weighted. (In Gray code the number would be 4)

(also have a look at http://www.asic-world.com/digital/numbering3.html )

Answer 2

Okay so, it's that additional constant (3 in the case of XS3 code and 4 in the case of Gray code, for example) being added to the initial sum that makes the entire number not be weighted? Or no?

Answer 3

Yes. The fact that you cannot simply take the sum of the weights, but that you have to do something different makes it non-weighted.

(btw, Gray code is a bit more complex that just adding +4. What I meant was that 0110 is 4 in Gray code)

Answer 4

Thanks a lot! :)